0.9
3.32/1.66+18.4 is 20.4
3.32 rounds down to 3
1.66 rounds up to 2
18.4 rounds down to 18
3/2+18 is 19.5
Since they want a positive difference the bigger number goes first
20.4-19.5=0.9
The answer for the first question: C The rotated image may be larger than the original image.
You will have infinite solutions
I'm going to assume that you meant 450kg for the combined weight, 190kg more and 3 Llamas. I'm pretty sure Llamas and Okapis don't weigh 450450450kg (that's 993,073,252 pounds). :)
x= Okapi weight
y= Llama weight
EQUATIONS:
There are 2 equations to be written:
1) 450kg is equal to the weight of one Okapi and one Llama
450kg= x + y
2) The weight of 3 llamas is equal to the weight of one Okapi plus 190kg.
3y=190kg + x
STEP 1:
Solve for one variable in one equation and substitute the answer in the other equation.
450kg= x + y
Subtract y from both sides
450-y =x
STEP 2:
Substitute (450-y) in second equation in place of x to solve for y.
3y=190kg + x
3y=190 + (450-y)
3y=640 -y
add y to both sides
4y=640
divide both sides by 4
y=160kg Llama weight
STEP 3:
Substitute 160kg in either equation to solve for x.
3y=190kg + x
3(160)=190 + x
480=190 + x
Subtract both sides by 190
290= x
x= 290kg Okapi weight
CHECK:
3y=190kg + x
3(160)=190 + 290
480=480
Hope this helps! :)
Work the information to set inequalities that represent each condition or restriction.
2) Name the
variables.
c: number of color copies
b: number of black-and-white copies
3)
Model each restriction:
i) <span>It
takes 3 minutes to print a color copy and 1 minute to print a
black-and-white copy.
</span><span>
</span><span>
3c + b</span><span>
</span><span>
</span><span>ii) He needs to print
at least 6 copies ⇒
c + b ≥ 6</span><span>
</span><span>
</span><span>iv) And must have
the copies completed in
no more than 12 minutes ⇒</span>
3c + b ≤ 12<span />
4) Additional restrictions are
c ≥ 0, and
b ≥ 0 (i.e.
only positive values for the number of each kind of copies are acceptable)
5) This is how you
graph that:
i) 3c + b ≤ 12: draw the line 3c + b = 12 and shade the region up and to the right of the line.
ii) c + b ≥ 6: draw the line c + b = 6 and shade the region down and to the left of the line.
iii) since c ≥ 0 and b ≥ 0, the region is in the
first quadrant.
iv) The final region is the
intersection of the above mentioned shaded regions.v) You can see such graph in the attached figure.